Results 1 to 9 of 9

Math Help - Help with nonlinear equation - reduction of order

  1. #1
    Junior Member
    Joined
    Jul 2011
    Posts
    26

    Help with nonlinear equation - reduction of order

    Using one of the methods of reduction of order, find the general solution of the following nonlinear equation:
    (y^3)y''=1

    So far, using a substitution of p=y', I have managed to do:
    (y^3)p(dp/dy)=1
    p dp = 1/(y^3) dy
    (p^2)/2 = -1/(2y^2) + c
    p^2 = -1/(y^2) + c

    Can someone please confirm if my working so far is correct or incorrect and where to go from here.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Help with nonlinear equation - reduction of order

    Quote Originally Posted by dwally89 View Post
    Using one of the methods of reduction of order, find the general solution of the following nonlinear equation:
    (y^3)y''=1

    So far, using a substitution of p=y', I have managed to do:
    (y^3)p(dp/dy)=1
    p dp = 1/(y^3) dy
    (p^2)/2 = -1/(2y^2) + c
    p^2 = -1/(y^2) + c

    Can someone please confirm if my working so far is correct or incorrect and where to go from here.

    Thanks
    So far so good...


    p=\sqrt{c -\frac{1}{y^2} }

    p=\frac{dy}{dx}

    \frac{dy}{dx}=\sqrt{c -\frac{1}{y^2} }

     dx=\frac{dy}{\sqrt{c-\frac{1}{y^2}}}

    Now take integral on both sides.
    Last edited by Also sprach Zarathustra; July 4th 2011 at 04:19 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2011
    Posts
    26

    Re: Help with nonlinear equation - reduction of order

    Quote Originally Posted by Also sprach Zarathustra View Post
    So far so good...


    p=\sqrt{c -\frac{1}{y^2} }

    p=\frac{dy}{dx}

    \frac{dy}{dx}=\sqrt{c -\frac{1}{y^2} }

     dx=\frac{dy}{\sqrt{c-\frac{1}{y^2}}}

    Now take integral on both sides.
    I managed to get to that step, but am not sure how to take the integral of the right hand side.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Help with nonlinear equation - reduction of order

    Quote Originally Posted by dwally89 View Post
    I managed to get to that step, but am not sure how to take the integral of the right hand side.

    Thanks
    \int\frac{dy}{\sqrt{c-\frac{1}{y^2}}}=\int \frac{y}{\sqrt{cy^2-1}}dy

    Now, substitute u=cy^2-1 then  du=2cy

    Try to finish...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2011
    Posts
    26

    Re: Help with nonlinear equation - reduction of order

    Quote Originally Posted by Also sprach Zarathustra View Post
    \int\frac{dy}{\sqrt{c-\frac{1}{y^2}}}=\int \frac{y}{\sqrt{cy^2-1}}dy

    Now, substitute u=cy^2-1 then  du=2cy

    Try to finish...
    So:
    y=\sqrt{cx^2+\frac{1}{c}}

    which then working backwards, to check that it works gives:
    y^3=({cx^2+\frac{1}{c}})^{1.5}
    y''=({cx^2+\frac{1}{c}})^{-1.5}
    y^3y''=1

    Right?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Help with nonlinear equation - reduction of order

    Quote Originally Posted by dwally89 View Post
    So:
    y=\sqrt{cx^2+\frac{1}{c}}

    which then working backwards, to check that it works gives:
    y^3=({cx^2+\frac{1}{c}})^{1.5}
    y''=({cx^2+\frac{1}{c}})^{-1.5}
    y^3y''=1

    Right?

    I don't think so...



    Steps after taking integration on both sides:

    \frac{y\sqrt{c-\frac{1}{y^2}}}{c}+A=x+B

    Or:

    y\sqrt{c-\frac{1}{y^2}}=x+K


    Or:

    \sqrt{cy^2-1}=x+K

    Or:


    cy^2-1=(x+K)^2


    Now solve for y.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jul 2011
    Posts
    26

    Re: Help with nonlinear equation - reduction of order

    Quote Originally Posted by Also sprach Zarathustra View Post
    I don't think so...



    Steps after taking integration on both sides:

    \frac{y\sqrt{c-\frac{1}{y^2}}}{c}+A=x+B

    Or:

    y\sqrt{c-\frac{1}{y^2}}=x+K


    Or:

    \sqrt{cy^2-1}=x+K

    Or:


    cy^2-1=(x+K)^2


    Now solve for y.
    I have a bad habit of forgetting to add constants when integrating...

    y=\sqrt{\frac{(x+K)^2+1}{c}}
    Is that right?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Help with nonlinear equation - reduction of order

    Quote Originally Posted by dwally89 View Post
    y=\sqrt{\frac{(x+K)^2+1}{c}}
    Is that right?
    y=\sqrt{\frac{(x+K)^2+1}{c}} or y=-\sqrt{\frac{(x+K)^2+1}{c}}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jul 2011
    Posts
    26

    Re: Help with nonlinear equation - reduction of order

    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Second order nonlinear nonhomogeneous differential equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: November 7th 2011, 04:51 AM
  2. First-order Nonlinear Equation with Linear Coefficients
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: September 13th 2011, 02:53 AM
  3. First-order nonlinear ordinary differential equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: September 12th 2011, 09:20 PM
  4. First-order Nonlinear Equation
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 11th 2011, 06:44 AM
  5. [SOLVED] Nonlinear first order equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: July 13th 2010, 05:48 AM

Search Tags


/mathhelpforum @mathhelpforum